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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: 5x-1/3x+2 = 5x+7/3x-1 3x+2/3x-2 – 6/2+3x = 9×2/9×2-4

Toán Lớp 8: 5x-1/3x+2 = 5x+7/3x-1
3x+2/3x-2 – 6/2+3x = 9×2/9×2-4

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    (5x-1)/(3x+2) = (5x+7)/(3x-1)
    ĐKXĐ : x \ne -2/3 , x \ne 1/3
    ⇔ (5x-1)(3x-1) = (3x+2)(5x+7)
    ⇔ 15x^2 – 8x + 1 = 15x^2 +31x + 14
    ⇔ 15x^2 – 8x = 15x^2 + 31x + 13
    ⇔ -8x = 31x + 13
    ⇔ -39x = 13
    ⇔ x = -1/3 (Thỏa mãn điều kiện)
    Vậy x \in {-1/3}
    (3x+2)/(3x-2) – 6/(2+3x) = (9x^2)/(9x^2-4)
    ĐKXĐ : x \ne \pm3/2
    ⇔ ((3x+2)(3x+2))/((3x-2)(3x+2)) – (6(3x-2))/((3x-2)(3x+2)) = (9x^2)/((3x-2)(3x+2))
    ⇒ (3x+2)^2 – 6(3x-2) = 9x^2
    ⇔ 9x^2 – 6x + 16 = 9x^2
    ⇔ -6x + 16 = 0
    ⇔ -6x = -16
    ⇔ x = 8/3 (Thỏa mãn điều kiện)
    Vậy x \in {8/3}

  2.  a)
    $\dfrac{5x-1}{3x+2}=\dfrac{5x+7}{3x-1}$
    ĐKXĐ: xne-2/3;xne1/3
    $⇔\dfrac{(5x-1)(3x-1)}{(3x+2)(3x-1)}=\dfrac{(3x+2)(5x+7)}{(3x+2)(3x-1)}$
    ⇒15x^2-5x-3x+1=15x^2+21x+10x+14
    ⇔15x^2-5x-3x-21x-15x^2-10x=14-1
    ⇔-39x=13
    ⇔x=-1/3(N)
    Vậy S={-1/3}
    b)
    \frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4} 
    ĐKXĐ: xne±2/3
    <=>\frac{(3x+2)^2}{(3x-2)(3x+2)}-\frac{6(3x-2)}{(3x-2)(2+3x)}=\frac{9x^2}{9x^2-4}
    =>9x^2+12x+4-18x+12=9x^2
    <=>9x^2+12x-18x-9x^2=-12-4
    <=>-6x=-16
    <=>x=8/3(N)
    Vậy S={8/3}

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222-9+11+12:2*14+14 = ? ( )